$I-AB$ be invertible $\Leftrightarrow$ $I-BA$ is invertible [duplicate]

assume $A,B\in M_n(F)$ if $I-AB$ be invertible then how to prove $I-BA$ is invertible and how find inverse of $I-BA$

marked as duplicate by Martin, Norbert, Cameron Buie, Amzoti, Davide GiraudoApr 18 '13 at 21:52

• See also: this and this among many others. – Martin Apr 18 '13 at 21:15

Hint :$(I - BA)(I + B(I - AB)^{-1}A) = I$

Hint:$$(I-BA)^{-1}=X$$ (say), Now expand left side. we get $$X=I+BA+ (BA)(BA)+(BA)(BA)(BA)+\dots$$ $$AXB=AB+(AB)^2+(AB)^3+(AB)^4+\dots$$ $$I+AXB=I+(AB)+(AB)^2+\dots+(AB)^n+\dots=(I-AB)^{-1}$$

Check yourself: $$(I+AXB)(I-AB)=I$$ and $$(I-AB)(I+AXB)=I$$

• It should be mentioned that the argument is heuristic. The expansion in the first line of proof is not really valid since it depends on a notion of limit. Of course the end result is that this heuristic argument does give the correct form of the inverse, which is then checked algebraically. – Ittay Weiss Apr 18 '13 at 21:05
• I like this way more than the top comment. Even though the top comment is meant to be a hint, I feel like it fully gives away the game while not explaining how it's done, while this answer actually is more helpful! – It'sNotALie. Apr 5 at 21:12

Claim: $\det(I-AB)=\det(I-BA)$.

Proof:

Case 1: $A$ is invertible. Exercise!

Case 2: $A$ is not invertible. Consider $k$ the algebraic closure of $F$.

Let $P(X)=\det(I-(A-xI)B)- \det(I-B(A-xI))$. By part Case 1, $P(x)=0$ for all $\{ x \in k| x \mbox{ is not an eigenvalue of} A \}$. Since $P(x)$ is a polynomial of degree $n$ which has infinitely many roots, it follows that $P(X) \equiv 0$, and hence $P(0)=0$.